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- 1970-1-1
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贴出来我的答案,求指教:
SQL部分:
- 1) SELECT SUM (CASE WHEN gender = 'male' THEN 1 ELSE 0 END) / COUNT(*)
- FROM customers
- 2) SELECT SUM(CASE WHEN product_id IS NULL THEN 0 ELSE 1 END) / COUNT(*) as ratio
- FROM(
- SELECT name, product_id FROM
- customers c.--
- LEFT JOIN sale s. 1point 3 acres
- ON c.id = s.customer_id
- GROUP BY c.id. Waral dи,
- ) temp ;. From 1point 3acres bbs
- 3) SELECT COUNT(DISTINCT(state))
- FROM stores;
- 4) SELECT count(*). From 1point 3acres bbs
- FROM
- (
- SELECT product_id
- FROM sale
- GROUP BY product_id
- HAVING SUM(unit_order) >= 5.--
- ) temp
- .google и
- 5) SELECT state,
- COUNT(id)
- FROM stores s
- WHERE area_squarefeet > 26000
- GROUP BY state
- 6) SELECT gender, MIN(birthday), MAX(birthday).--
- FROM customers
- GROUP BY gender
复制代码 . From 1point 3acres bbs
coding部分,用JAVA写的
- public static Map<Integer, Integer> counter(List<Integer> input) {
- Map<Integer, Integer> counter = new HashMap<>();
- for (int i : input) {
- counter.put(i, counter.containsKey(i) ? counter.get(i) + 1 : 1);
- }
- return counter;
- }
- /*Its actually asking for distinct*/
- public static List<Integer> distinctValues(List<Integer> input) {
- List<Integer> res = new LinkedList<>();
- for (int i : input) {
- if (!res.contains(i)) {
- res.add(i);
- }
- }
. check 1point3acres for more. - return res;
- }
- public static List<Integer> flatList(List<Object> input, List<Integer> output){
- for(Object i : input) {
- if (i instanceof List) {. Χ
- flatList((List<Object>) i, output);
- } else {
- output.add((int)i);.--
- }
- }
- return output; ..
- }. 1point3acres.com
- public static int percentagePos(List<Integer> input, double percentage) {
- return input.get((int) Math.ceil(input.size() * percentage));
- }
复制代码 |
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